121

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Number

$121$ (one hundred and twenty-one) is:

$11^2$


The $11$th square number after $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$:
$121 = 11 \times 11$


The $2$nd power of $11$ after $(1)$, $11$:
$121 = 11^2$


The $39$th semiprime:
$121 = 11 \times 11$


The second square number after $25$ of the form $n! + 1$:
$121 = 5! + 1 = 11^2$
where $!$ denotes the factorial function.


The $16$th powerful number after $1$, $4$, $8$, $9$, $16$, $25$, $27$, $32$, $36$, $49$, $64$, $72$, $81$, $100$, $108$


The $2$nd Fermat pseudoprime to base $3$ after $91$:
$3^{121} \equiv 3 \pmod {121}$


A palindromic number whose square is also palindromic:
$121^2 = 14641$


The $2$nd (with $4$) of the $2$ square numbers which are $4$ less than a cube:
$121 = 11^2 = 5^3 - 4$


The $6$th integer after $0$, $1$, $2$, $4$, $8$ which is palindromic in both decimal and ternary:
$121_{10} = 11 \, 111_3$


The $9$th positive integer which cannot be expressed as the sum of a square and a prime:
$1$, $10$, $25$, $34$, $58$, $64$, $85$, $91$, $121$, $\ldots$


The only square number which is the sum of consecutive powers of a positive integer:
$121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4$


The $7$th Smith number after $4$, $22$, $27$, $58$, $85$, $94$:
$1 + 2 + 1 = 1 + 1 + 1 + 1 = 4$


The $10$th square after $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$ which has no more than $2$ distinct digits


Also see


No further terms of this sequence are documented on $\mathsf{Pr} \infty \mathsf{fWiki}$.


Sources